What is the exact connection between Binomial coefficients and Factorials The factorial n! is the number of ways, the set {1, . . . , n} may be
ordered.
The binomial coefficient defines how many different ways there are to choose out of a group of n exactly k of them.
What is the connection between factorials and binomial coefficients?
 A: Here is the combinatorical approach:
Suppose that we choose $k$ out of $n$ elements where order does matter. We would have $n$ posibilities for the first choice, $n-1$ posibilities for the next and so on. So we have 
$$ n\cdot (n-1) \cdot (n-2)  ...  (n-k+1) = \frac{n!}{(n-k)!}$$
posibilities, but this is only when order does matter. When order doesn't matter, we would have to divide by the number of ways we can arrange the $k$ objects, which of course is $k!$. We thereby get that there is
$$\frac{n!}{k!(n-k)!}$$
ways to choose $k$ elements from a set of $n$ elements (when order does not matter).
A: The connection comes from the other definition of a binomial coefficient:
$\dbinom{n}{k}$ is the coefficient of $x^k$ in the expansion (by distributivity) of
$$(1+x)^n=\underbrace{(1+x)(1+x)\dotsm(1+x)}_{n\text{ factors}}$$
Indeed, to obtain $x^k$, in each of all factors $1+x$, we have to choose $k$ times $x$ and $n-k$ times $1$, in all possibles ways.
Using this approach, we can prove, differentiating $(1+x)^n$ and comparing the expansion of the derivative $n(1+x)^{n-1}$ and the derivative of the expansion, we obtain the following recursion formula:
$$\binom nk=\frac nk\binom{n-1}{k-1}$$
from which the usual formula for the value of $\binom nk$ in terms of factorials, is deduced.
